RD Sharma Class 11 Solutions Chapter - Brief Review of Cartesian System of Rectangular Coordinates | Mathematics (Maths) Class 11 - Commerce PDF Download

 Page 1


22. Brief Review of Cartesian System of Rectangular Co-
ordinates
Exercise 22.1
1. Question
If the line segment joining the points P(x
1
, y
1
) and Q(x
2
, y
2
) subtends an angle a at the origin O, 
prove that : OP. OQ cos a = x
1
 x
2
 + y
1
 y
2
.
Answer
Key points to solve the problem:
• Idea of distance formula- Distance between two points P(x
1
,y
1
) and Q(x
2
,y
2
) is given by- PQ = 
Given,
Two points P and Q subtends an angle a at the origin as shown in figure:
From figure we can see that points O,P and Q forms a triangle.
Clearly in ?OPQ we have:
 {from cosine formula in a triangle}
Page 2


22. Brief Review of Cartesian System of Rectangular Co-
ordinates
Exercise 22.1
1. Question
If the line segment joining the points P(x
1
, y
1
) and Q(x
2
, y
2
) subtends an angle a at the origin O, 
prove that : OP. OQ cos a = x
1
 x
2
 + y
1
 y
2
.
Answer
Key points to solve the problem:
• Idea of distance formula- Distance between two points P(x
1
,y
1
) and Q(x
2
,y
2
) is given by- PQ = 
Given,
Two points P and Q subtends an angle a at the origin as shown in figure:
From figure we can see that points O,P and Q forms a triangle.
Clearly in ?OPQ we have:
 {from cosine formula in a triangle}
?  …..equation 1
From distance formula we have-
OP = 
As, coordinates of O are (0, 0) ? x
2
 = 0 and y
2
 = 0
Coordinates of P are (x
1
, y
1
) ? x
1
 = x
1
 and y
1
 = y
1
= 
= 
Similarly, OQ = 
= 
And, PQ = 
? OP
2
 + OQ
2
 - PQ
2
 = 
? OP
2
 + OQ
2
 - PQ
2
 = 
Using (a-b)
2
 = a
2
 + b
2
 – 2ab
? OP
2
 + OQ
2
 - PQ
2
 = 2x
1
 x
2
 + 2y
1
 y
2
 ….equation 2
From equation 1 and 2 we have:
?  …Proved.
2. Question
The vertices of a triangle ABC are A(0, 0), B (2, -1) and C (9, 0). Find cos B.
Answer
Key points to solve the problem:
• Idea of distance formula- Distance between two points P(x
1
,y
1
) and Q(x
2
,y
2
) is given by- PQ = 
Given,
Coordinates of triangle and we need to find cos B which can be easily found using cosine formula.
See the figure:
Page 3


22. Brief Review of Cartesian System of Rectangular Co-
ordinates
Exercise 22.1
1. Question
If the line segment joining the points P(x
1
, y
1
) and Q(x
2
, y
2
) subtends an angle a at the origin O, 
prove that : OP. OQ cos a = x
1
 x
2
 + y
1
 y
2
.
Answer
Key points to solve the problem:
• Idea of distance formula- Distance between two points P(x
1
,y
1
) and Q(x
2
,y
2
) is given by- PQ = 
Given,
Two points P and Q subtends an angle a at the origin as shown in figure:
From figure we can see that points O,P and Q forms a triangle.
Clearly in ?OPQ we have:
 {from cosine formula in a triangle}
?  …..equation 1
From distance formula we have-
OP = 
As, coordinates of O are (0, 0) ? x
2
 = 0 and y
2
 = 0
Coordinates of P are (x
1
, y
1
) ? x
1
 = x
1
 and y
1
 = y
1
= 
= 
Similarly, OQ = 
= 
And, PQ = 
? OP
2
 + OQ
2
 - PQ
2
 = 
? OP
2
 + OQ
2
 - PQ
2
 = 
Using (a-b)
2
 = a
2
 + b
2
 – 2ab
? OP
2
 + OQ
2
 - PQ
2
 = 2x
1
 x
2
 + 2y
1
 y
2
 ….equation 2
From equation 1 and 2 we have:
?  …Proved.
2. Question
The vertices of a triangle ABC are A(0, 0), B (2, -1) and C (9, 0). Find cos B.
Answer
Key points to solve the problem:
• Idea of distance formula- Distance between two points P(x
1
,y
1
) and Q(x
2
,y
2
) is given by- PQ = 
Given,
Coordinates of triangle and we need to find cos B which can be easily found using cosine formula.
See the figure:
From cosine formula in ?ABC , We have:
cos B = 
using distance formula we have:
AB = 
BC = 
And, AC = 
? cos B = 
3. Question
Four points A (6, 3), B(-3, 5), C(4, -2) and D(x, 3x) are given in such a way that , find x.
Answer
Key points to solve the problem:
• Idea of distance formula- Distance between two points P(x
1
,y
1
) and Q(x
2
,y
2
) is given by- PQ = 
Page 4


22. Brief Review of Cartesian System of Rectangular Co-
ordinates
Exercise 22.1
1. Question
If the line segment joining the points P(x
1
, y
1
) and Q(x
2
, y
2
) subtends an angle a at the origin O, 
prove that : OP. OQ cos a = x
1
 x
2
 + y
1
 y
2
.
Answer
Key points to solve the problem:
• Idea of distance formula- Distance between two points P(x
1
,y
1
) and Q(x
2
,y
2
) is given by- PQ = 
Given,
Two points P and Q subtends an angle a at the origin as shown in figure:
From figure we can see that points O,P and Q forms a triangle.
Clearly in ?OPQ we have:
 {from cosine formula in a triangle}
?  …..equation 1
From distance formula we have-
OP = 
As, coordinates of O are (0, 0) ? x
2
 = 0 and y
2
 = 0
Coordinates of P are (x
1
, y
1
) ? x
1
 = x
1
 and y
1
 = y
1
= 
= 
Similarly, OQ = 
= 
And, PQ = 
? OP
2
 + OQ
2
 - PQ
2
 = 
? OP
2
 + OQ
2
 - PQ
2
 = 
Using (a-b)
2
 = a
2
 + b
2
 – 2ab
? OP
2
 + OQ
2
 - PQ
2
 = 2x
1
 x
2
 + 2y
1
 y
2
 ….equation 2
From equation 1 and 2 we have:
?  …Proved.
2. Question
The vertices of a triangle ABC are A(0, 0), B (2, -1) and C (9, 0). Find cos B.
Answer
Key points to solve the problem:
• Idea of distance formula- Distance between two points P(x
1
,y
1
) and Q(x
2
,y
2
) is given by- PQ = 
Given,
Coordinates of triangle and we need to find cos B which can be easily found using cosine formula.
See the figure:
From cosine formula in ?ABC , We have:
cos B = 
using distance formula we have:
AB = 
BC = 
And, AC = 
? cos B = 
3. Question
Four points A (6, 3), B(-3, 5), C(4, -2) and D(x, 3x) are given in such a way that , find x.
Answer
Key points to solve the problem:
• Idea of distance formula- Distance between two points P(x
1
,y
1
) and Q(x
2
,y
2
) is given by- PQ = 
• Area of a ?PQR – Let P(x
1,
y
1
) , Q(x
2
,y
2
) and R(x
3
,y
3
) be the 3 vertices of ?PQR.
Ar(?PQR) = 
Given, coordinates of triangle as shown in figure.
Also, 
ar(?DBC) = 
= 
Page 5


22. Brief Review of Cartesian System of Rectangular Co-
ordinates
Exercise 22.1
1. Question
If the line segment joining the points P(x
1
, y
1
) and Q(x
2
, y
2
) subtends an angle a at the origin O, 
prove that : OP. OQ cos a = x
1
 x
2
 + y
1
 y
2
.
Answer
Key points to solve the problem:
• Idea of distance formula- Distance between two points P(x
1
,y
1
) and Q(x
2
,y
2
) is given by- PQ = 
Given,
Two points P and Q subtends an angle a at the origin as shown in figure:
From figure we can see that points O,P and Q forms a triangle.
Clearly in ?OPQ we have:
 {from cosine formula in a triangle}
?  …..equation 1
From distance formula we have-
OP = 
As, coordinates of O are (0, 0) ? x
2
 = 0 and y
2
 = 0
Coordinates of P are (x
1
, y
1
) ? x
1
 = x
1
 and y
1
 = y
1
= 
= 
Similarly, OQ = 
= 
And, PQ = 
? OP
2
 + OQ
2
 - PQ
2
 = 
? OP
2
 + OQ
2
 - PQ
2
 = 
Using (a-b)
2
 = a
2
 + b
2
 – 2ab
? OP
2
 + OQ
2
 - PQ
2
 = 2x
1
 x
2
 + 2y
1
 y
2
 ….equation 2
From equation 1 and 2 we have:
?  …Proved.
2. Question
The vertices of a triangle ABC are A(0, 0), B (2, -1) and C (9, 0). Find cos B.
Answer
Key points to solve the problem:
• Idea of distance formula- Distance between two points P(x
1
,y
1
) and Q(x
2
,y
2
) is given by- PQ = 
Given,
Coordinates of triangle and we need to find cos B which can be easily found using cosine formula.
See the figure:
From cosine formula in ?ABC , We have:
cos B = 
using distance formula we have:
AB = 
BC = 
And, AC = 
? cos B = 
3. Question
Four points A (6, 3), B(-3, 5), C(4, -2) and D(x, 3x) are given in such a way that , find x.
Answer
Key points to solve the problem:
• Idea of distance formula- Distance between two points P(x
1
,y
1
) and Q(x
2
,y
2
) is given by- PQ = 
• Area of a ?PQR – Let P(x
1,
y
1
) , Q(x
2
,y
2
) and R(x
3
,y
3
) be the 3 vertices of ?PQR.
Ar(?PQR) = 
Given, coordinates of triangle as shown in figure.
Also, 
ar(?DBC) = 
= 
Similarly, ar(?ABC) = 
= 
? 
? 24.5 = 28x – 14
? 28x = 38.5
? x = 38.5/28 = 1.375
4. Question
The points A (2, 0), B(9, 1), C (11, 6) and D (4, 4) are the vertices of a quadrilateral ABCD. 
Determine whether ABCD is a rhombus or not.
Answer
Key points to solve the problem:
• Idea of distance formula- Distance between two points P(x
1
,y
1
) and Q(x
2
,y
2
) is given by- PQ = 
• Idea of Rhombus – It is a quadrilateral with all four sides equal.
Given, coordinates of 4 points that form a quadrilateral as shown in fig:
Using distance formula, we have:
AB =